OCHEM oxidizing quesiton....help!!

[potbelly]

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Hi all, I came up with one confusing oxidation concept.

In Destroyer Road Map 2 and ACS Ochem #5 on pg 138, both show the similar concepts...

Either Propyl benzene or Ethyl benzene can be oxidized all the way to Benzoic Acid when strond oxidizing reagents used such as KMnO4 or K2Cr2O7. Correct?

Then, here is my quesiton.

I attached a question which product is correct?
When we oxidize a primary alcohol, we simply change C-OH to COOH. Yet here, Primary alcohol is attached to benzyl group(ethyl benzene in this case) so I am confused if it is oxidized to benzoic acid....

Would you please cleaify this? Thanks!!

 

[klutzy1987]

Answer 1 is correct. Provided there is a hydrogen in the benzyllic position the entire chain will be oxidized and cut off and replaced with a cooh group forming benzoic acid.

 

[potbelly]

Answer 1 is correct. Provided there is a hydrogen in the benzyllic position the entire chain will be oxidized and cut off and replaced with a cooh group forming benzoic acid.

Thank you so much... Now it's clear🙂

BTW can anyone name the reactant compound? It is not ethyl benzyl alcohol, is it?

 

[prsndwg]

I think it is 3-phenyl propanol

 

[NeoAkira]

Answer 1 is correct. Provided there is a hydrogen in the benzyllic position the entire chain will be oxidized and cut off and replaced with a cooh group forming benzoic acid.

I disagree. Answer 1 would be correct if there was a double bond in the side chain. But since there isn't, only the hydroxyl carbon should be oxidized.

 

[fishstalker73]

answer 1 is correct.

 

[potbelly]

I disagree. Answer 1 would be correct if there was a double bond in the side chain. But since there isn't, only the hydroxyl carbon should be oxidized.

Can you explain little more detail on the double bond in the side chain?

 

[fishstalker73]

I disagree. Answer 1 would be correct if there was a double bond in the side chain. But since there isn't, only the hydroxyl carbon should be oxidized.

I think if there is a c=c bond on the side alkyl group, that alkyl group would turn into 1,2 diol instead.

 

[NeoAkira]

I think if there is a c=c bond on the side alkyl group, that alkyl group would turn into 1,2 diol instead.

Hmm, the problem is that it depends on the concentration of KMnO4 used. If it is concentrated I think a reaction much like this would happen:

http://en.wikipedia.org/wiki/Oxidation_of_primary_alcohols_to_carboxylic_acids (the first reaction)

As shown, there's no separation between the R group and the hydroxyl carbon.

As for what I was saying about a double bond in the side chain, if there was a double bond between the hydroxyl carbon and the carbon of the side chain bonded to the ring, then KMnO4 would break open the bond.

Much like the second reaction on this page: http://www.cliffsnotes.com/WileyCDA...ons.topicArticleId-22667,articleId-22625.html


Anyways, that's just my reasoning.

 

[fishstalker73]

This reaction involves a benzene ring, a different rule applied.

 

[Ryltar]

I don't remember the mechanism for this one, and you probably don't need to know it. I just remember oxidization of a benzene ring with a multi carbon chain substituent by a strong oxidizing agent such as KMnO4 or K2Cr2O7 will cleave down to the benzyl carbon forming benzylic acid.

 

[miedvied]

Oxidizing reactions on an alkyl chain attached to a benzene ring will, if there is a benzyllic hydrogen available, burn down until all that remains is a benzoic acid - regardless of the length of the starting alkyl chain.

 

[potbelly]

THank you guys!🙂